 So, today we will be talking about the effect of a substrate on the rate of an SN2 reaction. But, before we begin, let's recall the mechanism of an SN2 reaction. In this reaction, let's say X is the leaving group, so the incoming nucleophile Y minus attacks and X minus leaves simultaneously. Oh, it's a single step reaction and since it's a single step reaction, there's no intermediate fault. So how do I check what the rate depends on? Well, the transition state. The state where the incoming nucleophile is attacking and the leaving group is leaving simultaneously. A bond is being made between C and Y while a bond breaks between C and X. So, what are the factors on which the rate would depend? Stability of X minus, the more stable the leaving group, the quicker it would leave and the faster would be the rate. What are the factors? The concentration of the attacking nucleophile and its strength as well as the substrate. Today, we will be focusing on this substrate. So, we will keep all the other factors same for every reaction and we will compare the rates based on different substrates. Let's begin. Let's say we are asked to compare the rates of these two reactions which occur via SN2 mechanism. Wait, before we begin, what is DMSO? DMSO is dimethyl sulfoxide. It's a polar aprotic solvent. Okay, let's proceed now. What do I see here? The leaving group in both the cases is the same. Cl minus is going to leave and the attacking nucleophile is CN minus in either case, focusing on the substrate now. In the first case, a 1 degree carbon is attached to the leaving group while in the second case, a 2 degree carbon is attached to the leaving group. Remember, the nucleophile attacks and the leaving group leaves in a single step. So, why am I focusing on this 1 degree and 2 degree thingy? I'm gonna tell you in a bit. These are the two substrates which are written out fully spread out and the nucleophile CN minus has to attack now. It wouldn't want to attack from the same side where Cl is. Why? Because both are electron dense. They'll repel. So, it wouldn't want to take that chance. Where does it find its way now? It cannot come from a direction where this bulky group is because there would be steric hindrance. So, it has to find its way through some other path. What about the second case? Can't attack from the side where the leaving group is and there are two bulky groups. Oh, it's gonna be difficult for the nucleophile to find a way and attack the carbon and make the leaving group leave. So, there is more steric hindrance here. Let's also look at it from a transition state perspective. The nucleophile attacks, the leaving group leaves, the transition state has a lot of negative, negative charge lying around. These three groups, if they're bulky, the transition state will not be stable. They would all be fighting for space and if they are not so bulky, it wouldn't be much of a problem. If I look at this energy diagram, this is my reaction. It's a single step reaction. Here at the peak, there's transition state. Now, I want to increase the rate of the reaction. For that, I need to stabilize this transition state. More the bulky groups attached, more would be the repulsion, lesser would be the stability and more would be the energy. Not happening in our favor. But if the attached groups are less bulky, then the nucleophile will have easier pathway to attack the carbon. The transition state would also be much stable as there would be lesser repulsions happening and therefore the rate would be faster. So the basic idea is to stabilize the transition state so that the reaction occurs quickly. More the bulky groups attached on this carbon that is being attacked, less stable would be the transitions and lesser would be the rate. So when I compare the rates of a 1 degree, 2 degree and 3 degree substrate towards an SN2 reaction, I see the rate of a 1 degree substrate is faster than that of a 2 degree one, which is faster than that of a 3 degree one. This is exactly opposite of what we saw for SN1 reaction. Let's come back to our own question. Shall we? Which would react faster via SN2? The first one, right? Perfect. Let's try another question. We have to compare the reactivities of these two substrates towards SN2 reaction. The leaving group is still the same in each case, right? What about the bond between the carbon and the leaving group? We all know that chlorine has lone pair of electrons, right? The lone pair of electrons alternate to the pi bond. Does it ring a bell? Yes, resonance. There's a possibility of resonance in the first case and what is the result of this resonance? There's a partial double bond character between the carbon and the leaving group that is chlorine. What do I know about single bonds and double bonds? Well, double bonds are stronger than the single bonds for starters. They are more difficult to break. Yeah, the rate would be slower. But before I come to a conclusion, let me just find out if there's a possibility of resonance in the second case also. The lone pair are not alternate to the pi bond. No conjugation here. So the second one would react faster than the first one. But, but, but, would the first one want to react? I don't think so. It's very difficult for this bond to break. It wouldn't break easily. So vinylic halides or vinylic substrates do not undergo nucleophilic substitution reactions via SN2 pathway. Let's take up another problem. We have to compare the reactivities of these three substrates towards an SN2 reaction. Why don't you try this problem yourself before we do it together? Look carefully at the third one. Lone pair on the chlorine atom alternate to the pi bond would be involved in resonance. So there is resonance in the third case. There is a partial double bond character in the carbon-chlorine bond and therefore the third one would be non-reactive towards an SN2 reaction. Perfect. What about the first and the second? In either case, a one-degree carbon is attached to the leaving group. What do I compare now? Well, try drawing the transition state for each case and think which of the two transition states would be most stable. Do it and then we'll do it together. In both the first and the second cases, the carbon to which the leaving group is attached is attached to two hydrogen. What's different is the third thing. In the first case, it's CH2-CH3 while in the second case, it's CH double bond CH2, right? We saw before how the transition state was to electron dense. If I wanted to stabilize the transition state, I would want to stabilize this negative charge, right? That's what this CH double bond CH2 group does in the second case. It's an electron withdrawing group. It has a minus I effect. It is going to stabilize the transition state and the more stable the transition state, the faster would be the reaction. Let's take up one more problem now. So we are asked to compare the reactivities of these three substrates towards an SN2 reaction. Try it yourself and then we'll do it together. Focus on the first one. Here, the chlorine atom is directly attached to the benzene ring. Loan pair on the chlorine atom, alternate to the pi bond present in the benzene ring. You guessed it right. There's resonance. And since there's resonance, there's a partial double bond character in the CCL bond. Would it want to break? Not so much. And therefore, phenyl substrates or phenyl halides, just like the vinylic ones, wouldn't undergo nucleophilic substitution reaction via SN2 mechanism. So one wouldn't react. What about the other two? In both these cases, a one degree carbon is attached to the leaving ring. Right? What else? What do we do now? We draw the transition states. Let's do it. In the first case, there's a benzene ring. While in the second case, there's a cyclohexane ring present. Which of the two would be able to stabilize the transition state? First one because it has an electron withdrawing effect minus IA effect. Since the transition state is most stable, the second substrate would react the first test and the first one wouldn't want to react at all.